Question: Is ${927094}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {927094}= &&{9}\cdot100000+ \\&&{2}\cdot10000+ \\&&{7}\cdot1000+ \\&&{0}\cdot100+ \\&&{9}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {927094}= &&{9}(99999+1)+ \\&&{2}(9999+1)+ \\&&{7}(999+1)+ \\&&{0}(99+1)+ \\&&{9}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {927094}= &&\gray{9\cdot99999}+ \\&&\gray{2\cdot9999}+ \\&&\gray{7\cdot999}+ \\&&\gray{0\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {9}+{2}+{7}+{0}+{9}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${927094}$ is divisible by $9$ if ${ 9}+{2}+{7}+{0}+{9}+{4}$ is divisible by $9$ Add the digits of ${927094}$ $ {9}+{2}+{7}+{0}+{9}+{4} = {31} $ If ${31}$ is divisible by $9$ , then ${927094}$ must also be divisible by $9$ ${31}$ is not divisible by $9$, therefore ${927094}$ must not be divisible by $9$.